rainbow@ihuxe.UUCP (08/24/83)
The fact that you found a gold coin in the bottom drawer does not change
the problem in any way. Since the coins are distributed randomly into
the drawers, picking the bottom drawer is equivalent to picking any drawer.
Hence the probability is still 2/3 that the top drawer(other drawer) has
a gold coin in it.
To avoid senseless argument, I present the following example:
Lets say there are six cabinets. The distribtion of the coins are:
1)top=g bot=g
2)top=g bot=g
3)top=s bot=s
4)top=s bot=s
5)top=g bot=s
6)top=s bot=g
Now you open a bottom drawer and find a gold coin. 2 out of 3 times
the situation will now be case 1 or case 2. 1 out 3 times the
situation will now be case 6. Since each of the above cases has an equal
chance of occuring, the probabilites can be restated as follows:
1/3 top=g bot=g
1/3 top=s bot=s
1/6 top=g bot=s
1/6 top=s bot=g
Once you know that the bottom drawer has a gold coin, its twice as
likely that the top drawer will be gold rather than silver.
This example can also be used for the first statistics problem. The gold
coin chosen can be any one of six. 4 of these have gold coins have
gold coins in their companion drawer while 2 have silver.
I hope the gold coin problem is now a dead issue. The answer is obviously
2/3 in both problems. I'm getting tired of seeing silly attempts at justifying
the answer of 1/2. The problem is not even ambiguous in any way(except to
lawyers, that is, who manage to permutate everything to whatever suits
their needs). If on the other hand serious questions were still being
brought up, that would be another matter. But the responses have deteriorated
into words without sense.